A particle is subjected to two simple harmonic motions as : x1 = 7 5t cm and x2 = 2 7 (5t + 3 )cm where x is displacement and t is time in seconds. The maximum acceleration of the particle is x 10-2 m s-2. The value of x is :
- 5 7
- 125
- 25 7
- 175
Answer
(4) 175
Solution
Same = 5 rad s-1, amplitudes A1 = 7 cm, A2 = 2 7 cm, phase difference = 3
Resultant amplitude: A2 = A12 + A22 + 2A1A2
A2 = 7 + 28 + 2( 7)(2 7) ( 12 ) = 49
A = 7 cm
amax = 2 A = 25 7 = 175 cm s-2
amax = 175 10-2 m s-2 x = 175
Answer: (4)